\(\int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx\) [536]

Optimal result

Integrand size = 19, antiderivative size = 58 \[ \int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx=\frac {3 (a+b x)^{11/3} (c+d x)^{2/3} \operatorname {Hypergeometric2F1}\left (1,\frac {13}{3},\frac {14}{3},-\frac {d (a+b x)}{b c-a d}\right )}{11 (b c-a d)} \] Output:

3*(b*x+a)^(11/3)*(d*x+c)^(2/3)*hypergeom([1, 13/3],[14/3],-d*(b*x+a)/(-a*d 
+b*c))/(-11*a*d+11*b*c)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx=\frac {3 (a+b x)^{11/3} \sqrt [3]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {11}{3},\frac {14}{3},\frac {d (a+b x)}{-b c+a d}\right )}{11 b \sqrt [3]{c+d x}} \] Input:

Integrate[(a + b*x)^(8/3)/(c + d*x)^(1/3),x]
 

Output:

(3*(a + b*x)^(11/3)*((b*(c + d*x))/(b*c - a*d))^(1/3)*Hypergeometric2F1[1/ 
3, 11/3, 14/3, (d*(a + b*x))/(-(b*c) + a*d)])/(11*b*(c + d*x)^(1/3))
 

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {80, 79}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {\sqrt [3]{\frac {b (c+d x)}{b c-a d}} \int \frac {(a+b x)^{8/3}}{\sqrt [3]{\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}}}dx}{\sqrt [3]{c+d x}}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {3 (a+b x)^{11/3} \sqrt [3]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {11}{3},\frac {14}{3},-\frac {d (a+b x)}{b c-a d}\right )}{11 b \sqrt [3]{c+d x}}\)

Input:

Int[(a + b*x)^(8/3)/(c + d*x)^(1/3),x]
 

Output:

(3*(a + b*x)^(11/3)*((b*(c + d*x))/(b*c - a*d))^(1/3)*Hypergeometric2F1[1/ 
3, 11/3, 14/3, -((d*(a + b*x))/(b*c - a*d))])/(11*b*(c + d*x)^(1/3))
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

Maple [F]

\[\int \frac {\left (b x +a \right )^{\frac {8}{3}}}{\left (x d +c \right )^{\frac {1}{3}}}d x\]

Input:

int((b*x+a)^(8/3)/(d*x+c)^(1/3),x)
 

Output:

int((b*x+a)^(8/3)/(d*x+c)^(1/3),x)
 

Fricas [F]

\[ \int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {8}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((b*x+a)^(8/3)/(d*x+c)^(1/3),x, algorithm="fricas")
 

Output:

integral((b^2*x^2 + 2*a*b*x + a^2)*(b*x + a)^(2/3)/(d*x + c)^(1/3), x)
 

Sympy [F]

\[ \int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx=\int \frac {\left (a + b x\right )^{\frac {8}{3}}}{\sqrt [3]{c + d x}}\, dx \] Input:

integrate((b*x+a)**(8/3)/(d*x+c)**(1/3),x)
 

Output:

Integral((a + b*x)**(8/3)/(c + d*x)**(1/3), x)
 

Maxima [F]

\[ \int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {8}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((b*x+a)^(8/3)/(d*x+c)^(1/3),x, algorithm="maxima")
 

Output:

integrate((b*x + a)^(8/3)/(d*x + c)^(1/3), x)
 

Giac [F]

\[ \int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {8}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((b*x+a)^(8/3)/(d*x+c)^(1/3),x, algorithm="giac")
 

Output:

integrate((b*x + a)^(8/3)/(d*x + c)^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx=\int \frac {{\left (a+b\,x\right )}^{8/3}}{{\left (c+d\,x\right )}^{1/3}} \,d x \] Input:

int((a + b*x)^(8/3)/(c + d*x)^(1/3),x)
 

Output:

int((a + b*x)^(8/3)/(c + d*x)^(1/3), x)
 

Reduce [F]

\[ \int \frac {(a+b x)^{8/3}}{\sqrt [3]{c+d x}} \, dx=\left (\int \frac {\left (b x +a \right )^{\frac {2}{3}}}{\left (d x +c \right )^{\frac {1}{3}}}d x \right ) a^{2}+\left (\int \frac {\left (b x +a \right )^{\frac {2}{3}} x^{2}}{\left (d x +c \right )^{\frac {1}{3}}}d x \right ) b^{2}+2 \left (\int \frac {\left (b x +a \right )^{\frac {2}{3}} x}{\left (d x +c \right )^{\frac {1}{3}}}d x \right ) a b \] Input:

int((b*x+a)^(8/3)/(d*x+c)^(1/3),x)
 

Output:

int((a + b*x)**(2/3)/(c + d*x)**(1/3),x)*a**2 + int(((a + b*x)**(2/3)*x**2 
)/(c + d*x)**(1/3),x)*b**2 + 2*int(((a + b*x)**(2/3)*x)/(c + d*x)**(1/3),x 
)*a*b